Question

Answer 1

\(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)\)

\(=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2\)

\(=-x-15\)

Answer 2

a, \(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)=\dfrac{1}{2}x\left(x^2-2x+3\right)-5\left(x^2-2x+3\right)=\dfrac{1}{2}x^3-x^2+\dfrac{3}{2}x-5x^2+10x-15=\dfrac{1}{2}x^3-6x^2+\dfrac{23}{2}x-15\)

b, \(\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)=x^3-5x+3x^2-15+x^2+4x-x^3-4x^2=-x-15\)